Is ${329341}$ divisible by $9$ ?
Explanation: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {329341}= &&{3}\cdot100000+ \\&&{2}\cdot10000+ \\&&{9}\cdot1000+ \\&&{3}\cdot100+ \\&&{4}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {329341}= &&{3}(99999+1)+ \\&&{2}(9999+1)+ \\&&{9}(999+1)+ \\&&{3}(99+1)+ \\&&{4}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {329341}= &&\gray{3\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {3}+{2}+{9}+{3}+{4}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${329341}$ is divisible by $9$ if ${ 3}+{2}+{9}+{3}+{4}+{1}$ is divisible by $9$ Add the digits of ${329341}$ $ {3}+{2}+{9}+{3}+{4}+{1} = {22} $ If ${22}$ is divisible by $9$ , then ${329341}$ must also be divisible by $9$ ${22}$ is not divisible by $9$, therefore ${329341}$ must not be divisible by $9$.